is_pointer Template in C++
Last Updated :
19 Nov, 2018
The std::is_pointer template of C++ STL is used to check whether the given type is pointer or not. It returns a boolean value showing the same.
Syntax:
template <class T > struct is_pointer;
Parameter: This template accepts a single parameter T (Trait class) to check whether T is a pointer or not.
Return Value: This template returns a boolean value as shown below:
- True: if the type is a pointer.
- False: if the type is a non-pointer.
Below programs illustrate the std::is_pointer template in C++ STL:
Program 1:
#include <iostream>
#include <type_traits>
using namespace std;
class GFG {
};
int main()
{
cout << boolalpha;
cout << "is_pointer:"
<< endl;
cout << "GFG: "
<< is_pointer<GFG>::value << '\n' ;
cout << "GFG*: "
<< is_pointer<GFG*>::value << '\n' ;
cout << "GFG&: "
<< is_pointer<GFG&>::value << '\n' ;
cout << "nullptr_t: "
<< is_pointer<nullptr_t>::value << '\n' ;
return 0;
}
|
Output:
is_pointer:
GFG: false
GFG*: true
GFG&: false
nullptr_t: false
Program 2:
#include <iostream>
#include <type_traits>
using namespace std;
int main()
{
cout << boolalpha;
cout << "is_pointer:"
<< endl;
cout << "int: "
<< is_pointer< int >::value << '\n' ;
cout << "int *: "
<< is_pointer< int *>::value << '\n' ;
cout << "int **: "
<< is_pointer< int **>::value << '\n' ;
cout << "int ***: "
<< is_pointer< int ***>::value << '\n' ;
return 0;
}
|
Output:
is_pointer:
int: false
int *: true
int **: true
int ***: true
Program 3:
#include <iostream>
#include <type_traits>
using namespace std;
int main()
{
cout << boolalpha;
cout << "is_pointer:" << endl;
cout << "int *&: "
<< is_pointer< int *&>::value << '\n' ;
cout << "int *[10]: "
<< is_pointer< int * [10]>::value << '\n' ;
cout << "float *: "
<< is_pointer< float *>::value << '\n' ;
cout << "int[10]:"
<< is_pointer< int [10]>::value << '\n' ;
return 0;
}
|
Output:
is_pointer:
int *&: false
int *[10]: false
float *: true
int[10]:false
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